Amines — OCR A-Level Chemistry

Amines are organic derivatives of ammonia in which one or more hydrogen atoms are replaced by alkyl or aryl groups. Their basicity arises because the nitrogen atom carries a lone pair that can accept a proton (H⁺), forming a positively charged alkylammonium or arylammonium ion. In aqueous solution the equilibrium RNH₂ + H₂O ⇌ RNH₃⁺ + OH⁻ produces hydroxide ions, so amines act as weak bases. Alkyl groups release electron density towards nitrogen and stabilise the protonated cation, so primary, secondary and tertiary aliphatic amines are generally stronger bases than ammonia, while phenylamine is weaker because the lone pair is delocalised into the aromatic ring.

An amine behaves as a Brønsted–Lowry base because the nitrogen atom has a lone pair that accepts a proton. When a primary amine such as methylamine reacts with dilute hydrochloric acid, the lone pair forms a dative bond to H⁺ and the product is the substituted ammonium salt, for example CH₃NH₂ + HCl → CH₃NH₃⁺Cl⁻. The salt is ionic, so it is soluble in water but insoluble in non-polar organic solvents. Adding aqueous sodium hydroxide to the salt reverses the reaction by removing the proton, regenerating the free amine; this is used to liberate phenylamine from its salt after reduction of nitrobenzene. The same proton-acceptance idea explains why amines turn damp red litmus paper blue and why their aqueous solutions have a pH above 7. Base strength depends on how available the lone pair is: alkyl groups increase availability, whereas delocalisation into an aromatic ring decreases it.

This heading introduces the preparation routes for amines that the specification requires. It is a lead-in clause, so it carries no separate chemistry to learn on its own; its content is the sub-clauses that follow, beginning with aliphatic amines made by substituting haloalkanes with excess ethanolic ammonia and amines. Treat the heading as a signpost: when a question asks how an amine is prepared, identify which named route the stem describes, then check the reagents, the solvent and the excess condition before choosing an answer. A useful method is to write the general substitution equation for a haloalkane plus ammonia and confirm that the halogen leaves and the nitrogen of ammonia bonds to the carbon chain.

Aliphatic amines are prepared by nucleophilic substitution: ammonia or an amine acts as the nucleophile and attacks the electron-poor carbon of a haloalkane, displacing the halide ion. Ethanol is the solvent because it dissolves both the haloalkane and the ammonia while keeping them in one phase. Using excess ammonia favours the primary amine and limits further substitution, because ammonia outnumbers the haloalkane and the newly formed amine is less likely to compete. A worked example is bromoethane plus excess ethanolic ammonia giving ethylamine and ammonium bromide. When an amine rather than ammonia is the nucleophile, a secondary or tertiary amine can form, so the choice of nucleophile and its excess controls the product.

Nitroarenes such as nitrobenzene can be reduced to aromatic amines such as phenylamine. The standard laboratory route uses tin metal with concentrated hydrochloric acid: the tin is oxidised and supplies electrons, the acid provides protons, and the nitro group is reduced. The initial product in strong acid is the protonated amine (anilinium-type ion), so the mixture is then treated with alkali, typically sodium hydroxide, to liberate the free aromatic amine, which can be separated by steam distillation. Learners should track oxidation states: the nitrogen in the nitro group is reduced, while tin is oxidised from 0 to +2 (as Sn²⁺) or +4 (as Sn⁴⁺). Recognising which species is oxidised and which is reduced, and why the alkaline step is needed, is the core of this item.